the questions in level 1 are introductory problems. the hints contain links to videos covering this content…

Explanation:

Step1: Split the integral

According to the property of definite - integral (\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx) ((a < c < b)), we split (\int_{-\pi}^{\pi}f(x)dx) into (\int_{-\pi}^{0}f(x)dx+\int_{0}^{\pi}f(x)dx). Since (f(x)=\begin{cases}2x^{4},&-\pi\leq x < 0\5\sin(x),&0\leq x\leq\pi\end{cases}), then (\int_{-\pi}^{\pi}f(x)dx=\int_{-\pi}^{0}2x^{4}dx+\int_{0}^{\pi}5\sin(x)dx).

Step2: Integrate (2x^{4})

Use the power - rule for integration (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). For (\int_{-\pi}^{0}2x^{4}dx), we have (2\int_{-\pi}^{0}x^{4}dx). [ \begin{align*} 2\int_{-\pi}^{0}x^{4}dx&=2\left[\frac{x^{5}}{5}\right]_{-\pi}^{0}\ &=2\left(0-\frac{(-\pi)^{5}}{5}\right)\ &=\frac{2\pi^{5}}{5} \end{align*} ]

Step3: Integrate (5\sin(x))

Use the integral formula (\int\sin(x)dx=-\cos(x)+C). For (\int_{0}^{\pi}5\sin(x)dx), we have (5\int_{0}^{\pi}\sin(x)dx). [ \begin{align*} 5\int_{0}^{\pi}\sin(x)dx&=5[-\cos(x)]_{0}^{\pi}\ &=5(-\cos(\pi)+\cos(0))\ &=5(-(-1)+1)\ &=10 \end{align*} ]

Step4: Sum the two results

(\int_{-\pi}^{\pi}f(x)dx=\frac{2\pi^{5}}{5}+10)

Answer:

(\frac{2}{5}\pi^{5}+10)